Problem: $ g(x) = \int_{\,1}^{\,x}\sqrt{2t+7}\,dt\,$ $ g\,^\prime(9)\, =$
Answer: The Fundamental Theorem of Calculus If $~ g(x)=\int_a^xf(t)\,dt\,$, then $~g^\prime (x)=f(x)\,$ This only works if $f$ is continuous on $[a,b]$. Thankfully, the function $f(t) = \sqrt{2t + 7}$ is continuous on $[1,9]$. Applying the theorem We're given: $ g(x) = \int_{1}^{x}\sqrt{2t+7}\,dt$ So the theorem tells us: $ g\,^\prime(x) =\sqrt{2x+7}$ Evaluating $g'(9)$ $ g\,^\prime(9) = \sqrt{2\cdot9+7}=\sqrt{25}=5$ The answer: $g'(9)=5$